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# Imaginary Unit and Polar Bijection

We often see real numbers on number lines and axes in the Cartesian coordinate system. These digits are clearly visible to the naked eye which then allows us to make indications of the points we intend to mark. Well, what if I told you that there exists another axis that we cannot explicitly see on a plane, but rather "imagine" it? In fact, a completely different type of number is found on this imaginary axis. These numbers are called imaginary numbers. With that in mind, it then raises the question: what does it mean for a number to be imaginary?

In this article, we aim to understand the concept of an imaginary unit, which is the building block of an imaginary number. Throughout this topic, we shall also familiarise ourselves with some of its notable properties and subsequently bridge this idea onto complex numbers and their polar form.

## What is the Imaginary Unit?

Before introducing ourselves to the imaginary unit, it may be helpful to recall the definition of a square root to ease us into the topic.

The square root of a number, n is a number such that m2 = n or in other words, a number m whose square is n. The square root symbol is denoted by √.

For example, 5 is a square root of 25 since 52 = 25. Further note that –5 is also a square root of 25 since (–5)2 = 25. Writing this in notation form, we obtain the following.

√25 equals 5 or –5

While this seems rather straightforward, the number within a square root symbol happens to always be positive. Try plugging the square root of any positive real number into your calculator. See that you will always obtain another positive real number in return. Now, try putting in the square root of a negative real number. You will find that your calculator will display an error message. So does this mean that we cannot acquire the square root negative real numbers?

Let us return to our previous example. Say we had √–25 instead. One possible product of –25 is –5 times 5. Since –5 does not equal 5, this means that we cannot express √–25 as the square root form in the definition above. Does that mean we can leave √–25 as is? Well, not exactly. To answer this question, we can indeed simplify this expression further using the imaginary unit. Let us define this new term as below.

The imaginary unit is denoted by the letter i and is equivalent to i = √–1. Thus, i 2 = –1.

In some textbooks, the imaginary unit is also referred to as the imaginary factor.

Essentially speaking, this means that the number i is the principal square root of –1. This brings us to the following definition.

A pure imaginary number stems from the square root of a negative real number.

6i, –11i and i√3 are examples of pure imaginary numbers. Now that we have established this pair of definitions, we can infer the following concept.

The imaginary unit, i is usually written before radical symbols to make it clear that it is not under the radical.

Interpreting Imaginary Numbers

For any positive real number β

$\sqrt{-{\beta }^{2}}=\sqrt{{\beta }^{2}}×\sqrt{-1}=\beta \mathbit{i}$.

Let us now observe how we might apply this idea in the following sections.

### Square Root of Negative Numbers

The square root of any negative number, say β, can be written as the form

$\sqrt{-\beta }=i\sqrt{\beta }$.

Here is a brief example.

Find the square root of –144.

Solution

$\sqrt{-144}=i\sqrt{144}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-144}=12i$

Find the square root of$-\frac{7}{9}$

Solution

$\sqrt{-\frac{7}{9}}=i\sqrt{\frac{7}{9}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-\frac{7}{9}}=i\frac{\sqrt{7}}{\sqrt{9}}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-\frac{7}{9}}=\frac{\sqrt{7}}{3}i$

### Square of Imaginary Numbers

The square of every imaginary number, say β, can be expressed as the form

${\left(\beta i\right)}^{2}=-{\beta }^{2}$.

Calculate the square of 17i.

Solution

${\left(17i\right)}^{2}={17}^{2}{i}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(17i\right)}^{2}=289\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒{\left(17i\right)}^{2}=-289$

Calculate the square of -3i.

Solution

${\left(-3i\right)}^{2}={\left(-3\right)}^{2}{i}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(-3i\right)}^{2}=9\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒{\left(-3i\right)}^{2}=-9$

### The Cycle of the Imaginary Unit

The imaginary unit cycles through 4 different values each time we multiply it by itself.

 Cycle 1 Cycle 2 Cycle 3 Cycle 4 $i=\sqrt{-1}$ ${i}^{2}=-1$ ${i}^{3}=-\sqrt{-1}$ ${i}^{4}=1$ ${i}^{5}=\sqrt{-1}$ ${i}^{6}=-1$ ${i}^{7}=-\sqrt{-1}$ ${i}^{8}=1$ Cycle continues...

This is a very important property you should familiarize yourself with as it may help solve equations involving complex roots of higher degrees.

### Products of Pure Imaginary Numbers

We can indeed apply the cumulative and associative properties of multiplication to pure imaginary numbers. Below are some worked examples.

Simplify $5i×3i$.

Solution

$5i×3i=\left(5×3\right)×\left(i×i\right)\phantom{\rule{0ex}{0ex}}⇒5i×3i=15{i}^{2}\phantom{\rule{0ex}{0ex}}⇒5i×3i=15\left(-1\right)\phantom{\rule{0ex}{0ex}}⇒5i×3i=-15$

Simplify $\sqrt{-17}×\sqrt{-2}$.

Solution

$\sqrt{-17}×\sqrt{-2}=i\sqrt{17}×i\sqrt{2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-17}×\sqrt{-2}=\left(i×i\right)×\sqrt{17×2}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-17}×\sqrt{-2}={i}^{2}×\sqrt{34}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-17}×\sqrt{-2}=\left(-1\right)×\sqrt{34}\phantom{\rule{0ex}{0ex}}⇒\sqrt{-17}×\sqrt{-2}=-\sqrt{34}$

Simplify i 31.

Solution

${i}^{31}={i}^{1}×{i}^{30}\phantom{\rule{0ex}{0ex}}⇒{i}^{31}=i×{\left({i}^{2}\right)}^{15}\phantom{\rule{0ex}{0ex}}⇒{i}^{31}=i×{\left(-1\right)}^{15}\phantom{\rule{0ex}{0ex}}⇒{i}^{31}=i×-1\phantom{\rule{0ex}{0ex}}⇒{i}^{31}=-i$

## Solving Equations Using the Imaginary Unit

The imaginary unit is also involved in solving quadratic equations. This makes use of the Square Root Property as stated below.

Square Root Property

For any real number n, if m2 = n, then m = ±√n .

Below are several worked examples showing this.

Solve x2 + 1 = 0.

Solution

${x}^{2}+1=0⇒{x}^{2}=-1⇒x=±\sqrt{-1}⇒x=±i$${x}^{2}+1=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-1\phantom{\rule{0ex}{0ex}}⇒x=±\sqrt{-1}\phantom{\rule{0ex}{0ex}}⇒x=±i$

Solve 25x2 + 16 = 0.

Solution

$25{x}^{2}+16=0\phantom{\rule{0ex}{0ex}}⇒25{x}^{2}=-16\phantom{\rule{0ex}{0ex}}⇒{x}^{2}=-\frac{16}{25}\phantom{\rule{0ex}{0ex}}⇒x=±\sqrt{-\frac{16}{25}}\phantom{\rule{0ex}{0ex}}⇒x=±i\sqrt{\frac{16}{25}}\phantom{\rule{0ex}{0ex}}⇒x=±i\frac{\sqrt{16}}{\sqrt{25}}\phantom{\rule{0ex}{0ex}}⇒x=±\frac{4}{5}i$

## The Imaginary Unit and Complex Numbers

In this segment, we shall observe a link between the imaginary unit and imaginary numbers. An imaginary number is also referred to as a complex number and is often denoted by the letter z. The complex number, z = a + bi is made up of the two components:

1. The Real Part, described by Re (z) = a;

2. The Imaginary Part, described by Im (z) = b.

For a complex number, z = a + bi:

• If Im (z) = 0, then z = a is a real number
• If Re (z) = 0, then z = bi is said to be purely imaginary

Let us now define the modulus and argument of a complex number.

### The Modulus and Argument

Say we have a complex number z = a + bi.

The modulus of a complex number is represented by

$\left|z\right|=\sqrt{{a}^{2}+{b}^{2}}$.

This may also be referred to as the absolute value of a complex number.

The argument of a complex number is defined by

$arg\left(z\right)={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$.

Now that we have these formulas secured, we shall now use them to sketch a given complex number on a complex plane called the Argand Diagram.

### The Argand Diagram

A complex number z = a + bi can be illustrated as a point on the Im(z) – Re(z) complex plane as shown below. This is called the Argand Diagram of a complex number.

The Argand Diagram, StudySmarter Originals

Essentially, a complex number z = a + bi can be represented as an ordered pair a and b (the point in pink). The length of the line segment is the modulus r = |z|, which by definition is the distance of number z from the origin. And the direction is determined by the angle θ = arg z. As the main topic of discussion here is the imaginary unit, we shall only touch on a brief explanation of complex numbers. A more detailed overview of this can be found here: Complex Numbers.

### Polar Form of Complex Numbers

In this section, we shall look at expressing complex numbers in polar form.

A complex number z = a + bi with coordinates (a, b) can be represented by its polar form given by

z = r cosθ + i r sinθ = r (cosθ + i sinθ)

where r is the modulus of z and θ is the argument of z.

We can abbreviate this polar form by z = re. The polar form of a complex number is also called Euler's Relation.

The polar form can also be referred to as the trigonometric form in some textbooks.

Additionally, r > 0 and –π < θ < π.

### Derivation of the Polar Form

How does a complex number convert to its polar form? Let us observe how this is derived. We shall now refer to our previous graph showing a typical complex number on an Argand Diagram.

Derivation of polar form, StudySmarter Originals

Using trigonometric ratios, we obtain the following.

$\mathrm{cos}\theta =\frac{a}{r}\mathbit{a}\mathbit{n}\mathbit{d}\mathrm{sin}\theta =\frac{b}{r}$

This yields,

$a=r\mathrm{cos}\theta \mathbit{a}\mathbit{n}\mathbit{d}b=r\mathrm{sin}\theta$

From the definition of a complex number, we have that

$z=a+bi\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒z=r\mathrm{cos}\theta +ir\mathrm{sin}\theta \phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒z=r\left(\mathrm{cos}\theta +i\mathrm{sin}\theta \right)$

Now, referring to the image, the real and imaginary parts of the number z form a right-angle triangle. With that, we can apply none other than Pythagoras' theorem! We then reach the following derivation.

${r}^{2}={a}^{2}+{b}^{2}\mathbit{a}\mathbit{n}\mathbit{d}\mathrm{tan}\theta =\frac{b}{a}$

This gives us,

$r=\sqrt{{a}^{2}+{b}^{2}}\mathbit{a}\mathbit{n}\mathbit{d}\theta ={\mathrm{tan}}^{-1}\left(\frac{b}{a}\right)$

Thus, by convention, the horizontal axis is the real axis and the vertical axis is the imaginary axis in an Argand diagram. Below are some useful concepts and formulas involving Euler's Relation.

Formulas Involving Euler's Relation

1. ${e}^{i\theta }=\mathrm{cos}\left(\theta \right)+i\mathrm{sin}\left(\theta \right)$
2. ${e}^{i\theta }+{e}^{-i\theta }=2\mathrm{cos}\theta$
3. ${e}^{i\theta }-{e}^{-i\theta }=2i\mathrm{sin}\theta$

Concepts Involving Euler's Relation

1. ${e}^{2n\pi i}=1$
2. ${e}^{\frac{\pi }{2}i}=i$
3. ${e}^{\pi i}=-1$
4. ${e}^{\frac{3\pi }{2}i}=-i$

The polar form of a complex number is helpful when dealing with the powers and roots of complex numbers. Writing complex numbers in their polar form allows us to simplify such complex numbers which will indeed make calculations easier to solve. You can explore this concept in more depth here: Polar Coordinates and Representation of Complex Numbers.

We shall end this discussion with some worked examples of representing complex numbers in their polar forms.

Write α = 3 + 4i in polar form.

Solution

The modulus is given by

$r=\sqrt{{3}^{2}+{4}^{2}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{25}\phantom{\rule{0ex}{0ex}}⇒r=5$

The argument is given by

$\theta ={\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)$

Thus, the polar form is

$\alpha =3+4i\phantom{\rule{0ex}{0ex}}⇒\alpha ={e}^{5i{\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)}$

or

$\alpha =3+4i\phantom{\rule{0ex}{0ex}}⇒\alpha =5\left[\mathrm{cos}\left({\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\right)+i\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(\frac{4}{3}\right)\right)\right]$

Express $\beta =\frac{5}{2}-\frac{1}{2}i$ in polar form.

Solution

The modulus is given by

$r=\sqrt{{\left(\frac{5}{2}\right)}^{2}+{\left(-\frac{1}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{\frac{26}{4}}\phantom{\rule{0ex}{0ex}}⇒r=\sqrt{\frac{13}{2}}$

The argument is given by

Thus, the polar form is

$\beta =\frac{5}{2}-\frac{1}{2}i\phantom{\rule{0ex}{0ex}}⇒\beta ={e}^{-i\sqrt{\frac{13}{2}}{\mathrm{tan}}^{-1}\left(5\right)}$

or

$\beta =\frac{5}{2}-\frac{1}{2}i\phantom{\rule{0ex}{0ex}}⇒\beta =\sqrt{\frac{13}{2}}\left[\mathrm{cos}\left({\mathrm{tan}}^{-1}\left(5\right)\right)-i\mathrm{sin}\left({\mathrm{tan}}^{-1}\left(5\right)\right)\right]$

## Imaginary Unit and Polar Bijection - Key takeaways

• The imaginary unit is i = √–1 ⇒ i2 = –1
• The imaginary unit can be used to solve polynomial equations.
• The imaginary unit has a cycle of 4.
• The polar form of a complex number z = a + bi is z = eirθ = r (cos θ + i sin θ)
• The polar form of a complex number is helpful when dealing with the powers and roots of complex numbers.

## Frequently Asked Questions about Imaginary Unit and Polar Bijection

For a complex number of the form z = a + ib, the polar form is denoted by z = eirθ where r is the modulus and θ is the argument

i2 = -1 or i = √-1

The complex conjugate function is bijective

The polar form of a complex number is helpful when dealing with the powers and roots of complex numbers

The imaginary unit of a complex number is i = √-1

## Final Imaginary Unit and Polar Bijection Quiz

Question

What do you get when you multiply a complex number $$z$$ and its complex conjugate $$\bar{z}$$ together?

$$|z|^2$$.

Show question

Question

What is the plane in which complex numbers are plotted known as?

Complex plane or Argand plane

Show question

Question

What is Modulus of a complex number?

The distance of a complex point from the origin on a complex plane is called the modulus of a complex number.

Show question

Question

How will the phase of a complex number z change when it is raised to a power of n?

The phase will now be n times the original phase.

Show question

Question

Solve (2 + 7i) + (3 − 4i)

5 + 3i

Show question

Question

Solve (6 + 9i) + (12 − 17i)

18 - 8i

Show question

Question

Calculate (9 + 5i) − (4 + 7i)

5 − 2i

Show question

Question

Calculate (2 + 7i) − (13 + 5i)

-11 + 2i

Show question

Question

Given a complex number z = 5 - 2i, find 6z

30 - 12i

Show question

Question

Given a complex number z = 9 + 3i, find -2z

-18 - 6i

Show question

Question

Evaluate (3 + 2i) (5 + 6i)

3 + 28i

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Question

Evaluate (5 - 9i) (4 + 2i)

38 -26 i

Show question

Question

What is the complex conjugate of -2 + i?

-2 - i

Show question

Question

What is the complex conjugate of 15 - 4i?

15 + 4i

Show question

Question

Express a complex number z in polar form

z = eirθ

Show question

Question

What do the r and θ represent in the polar form of complex numbers z = eirθ ?

r = modulus

θ = argument

Show question

Question

How do you write the polar form in terms of Euler's relation?

z = eirθ = r (cos θ + i sin θ)

Show question

Question

Why is the polar form useful when dealing with complex numbers?

The polar form of a complex number is helpful when powers and roots are involved in complex numbers

Show question

Question

What is the square of the imaginary unit equal to?

-1

Show question

Question

What is the polynomial representation when it comes to complex numbers?

The polynomial representation takes the form,

$z = a + bi$

where, $$z$$ is a complex number, $$a$$ and $$b$$ are real numbers, $$i = \sqrt{-1}$$.

Show question

Question

Express $$z=6+2i$$ in Cartesian form.

$$z=(6,2)$$

Show question

Question

Express $$z=7-3i$$ in Cartesian form.

$$z=(7,-3)$$

Show question

Question

Express $$z=-6-9i$$ in Cartesian form.

$$z=(-6,-9)$$

Show question

Question

The complex number $$z$$ in Cartesian representation is represented as $$z=(3,-4)$$. What is the length of the line representing $$z$$ in graphical form?

5

Show question

Question

The complex number $$z$$ in Cartesian representation is represented as $$z=(-12,9)$$. What is the length of the line representing $$z$$ in graphical form?

15

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Question

State whether the following statement is true or false:

Complex numbers can be represented on a coordinate plane.

False

Show question

Question

State whether the following statement is true or false:

Complex numbers are bi-dimensional.

True

Show question

Question

The complex number $$z$$ in Cartesian representation is represented as $$z=(-6,-8)$$. What is the length of the line representing $$z$$ in graphical form?

10

Show question

Question

Represent $$z=2+3i$$ using a matrix.

$z = \begin{pmatrix} 2 & -3 \\ 3 & 2 \end{pmatrix}$

Show question

Question

Represent $$z=-5+2i$$ using matrix representation.

$z = \begin{pmatrix} -5 & -2 \\ 2 & 5 \end{pmatrix}$

Show question

Question

Represent $$z=-1-7i$$ using matrix representation.

$z = \begin{pmatrix} -1 & 7 \\ -7 & -1 \end{pmatrix}$

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Question

Add the 2 complex numbers

$z_1 = \begin{pmatrix} 4 & -2 \\ 2 & 4 \end{pmatrix}$

and

$z_2 = \begin{pmatrix} 7 & -5 \\ 5& 7 \end{pmatrix}$

$\begin{pmatrix} 11 & -7 \\ 7 & 11 \end{pmatrix}$

Show question

Question

Given

$z_1 = \begin{pmatrix} 3 & -2 \\ 2 & 3 \end{pmatrix}$

and

$z_2 = \begin{pmatrix}1 & -5 \\ 5 & 1\end{pmatrix}$

Find $$z=z_1 - z_2$$.

\begin{pmatrix} 2& 3 \\ -3 & 2\end{pmatrix}

Show question

Question

Which of the following representations is

$$z=6+9i$$

Polynomial representation

Show question

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